# Modular Multiplicative Inverse

## Definition

A modular multiplicative inverse of an integer $a$ is an integer $x$ such that $a \cdot x$ is congruent to $1$ modular some modulus $m$. To write it in a formal way: we want to find an integer $x$ so that

$a \cdot x \equiv 1 \mod m.$

We will also denote $x$ simply with $a^{-1}$.

We should note that the modular inverse does not always exist. For example, let $m = 4$, $a = 2$. By checking all possible values modulo $m$ is should become clear that we cannot find $a^{-1}$ satisfying the above equation. It can be proven that the modular inverse exists if and only if $a$ and $m$ are relatively prime (i.e. $\gcd(a, m) = 1$).

In this article, we present two methods for finding the modular inverse in case it exists, and one method for finding the modular inverse for all numbers in linear time.

## Finding the Modular Inverse using Extended Euclidean algorithm

Consider the following equation (with unknown $x$ and $y$):

$a \cdot x + m \cdot y = 1$

This is a Linear Diophantine equation in two variables. As shown in the linked article, when $\gcd(a, m) = 1$, the equation has a solution which can be found using the extended Euclidean algorithm. Note that $\gcd(a, m) = 1$ is also the condition for the modular inverse to exist.

Now, if we take modulo $m$ of both sides, we can get rid of $m \cdot y$, and the equation becomes:

$a \cdot x \equiv 1 \mod m$

Thus, the modular inverse of $a$ is $x$.

The implementation is as follows:

int x, y;
int g = extended_euclidean(a, m, x, y);
if (g != 1) {
cout << "No solution!";
}
else {
x = (x % m + m) % m;
cout << x << endl;
}


Notice that we way we modify x. The resulting x from the extended Euclidean algorithm may be negative, so x % m might also be negative, and we first have to add m to make it positive.

## Finding the Modular Inverse using Binary Exponentiation

Another method for finding modular inverse is to use Euler's theorem, which states that the following congruence is true if $a$ and $m$ are relatively prime:

$a^{\phi (m)} \equiv 1 \mod m$

$\phi$ is Euler's Totient function. Again, note that $a$ and $m$ being relative prime was also the condition for the modular inverse to exist.

If $m$ is a prime number, this simplifies to Fermat's little theorem:

$a^{m - 1} \equiv 1 \mod m$

Multiply both sides of the above equations by $a^{-1}$, and we get:

• For an arbitrary (but coprime) modulus $m$: $a ^ {\phi (m) - 1} \equiv a ^{-1} \mod m$
• For a prime modulus $m$: $a ^ {m - 2} \equiv a ^ {-1} \mod m$

From these results, we can easily find the modular inverse using the binary exponentiation algorithm, which works in $O(\log m)$ time.

Even though this method is easier to understand than the method described in previous paragraph, in the case when $m$ is not a prime number, we need to calculate Euler phi function, which involves factorization of $m$, which might be very hard. If the prime factorization of $m$ is known, then the complexity of this method is $O(\log m)$.

## Finding the modular inverse for every number modulo $m$ {#mod-inv-all-num}

The problem is the following: we want to compute the modular inverse for every number in the range $[1, m-1]$.

Applying the algorithms described in the previous sections, we can obtain a solution with complexity $O(m \log m)$.

Here we present a better algorithm with complexity $O(m)$. However for this specific algorithm we require that the modulus $m$ is prime.

We denote by $\text{inv}[i]$ the modular inverse of $i$. Then for $i > 1$ the following equation is valid:

$\text{inv}[i] = - \left\lfloor \frac{m}{i} \right\rfloor \cdot \text{inv}[m \bmod i] \bmod m$

Thus the implementation is very simple:

inv = 1;
for(int i = 2; i < m; ++i)
inv[i] = m - (m/i) * inv[m%i] % m;


### Proof

We have:

$m \bmod i = m - \left\lfloor \frac{m}{i} \right\rfloor \cdot i$

Taking both sides modulo $m$ yields:

$m \bmod i \equiv - \left\lfloor \frac{m}{i} \right\rfloor \cdot i \mod m$

Multiply both sides by $i^{-1} \cdot (m \bmod i)^{-1}$ yields

$(m \bmod i) \cdot i^{-1} \cdot (m \bmod i)^{-1} \equiv -\left\lfloor \frac{m}{i} \right\rfloor \cdot i \cdot i^{-1} \cdot (m \bmod i)^{-1} \mod m,$

which simplifies to:

$i^{-1} \equiv -\left\lfloor \frac{m}{i} \right\rfloor \cdot (m \bmod i)^{-1} \mod m,$